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Question

A block of mass m is on an inclined plane of angle θ. The coefficient of friction between the block and the plane is μ and tanθ>μ. The block is held stationary by applying a force P parallel to the plane. The direction of force pointing up the plane is taken to be positive. As P is varied from P1=mg(sinθμcosθ) to P2=mg(sinθ+μcosθ), the frictional force f versus P graph will look like:

A
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B
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C
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D
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Solution

The correct option is A
As tanθ>μ, the block has a tendency to move down the incline.
At equilibrium P+f=mg sinθf=mg sinθP


Now as P increases the value of f decreases linearly with resepct to P.

When P=mg sinθ,f=0

When P is increased further, the block will have a tendency to move upwards along the inclined plane.
Therefore the frictional force acts downwards along the incline in this case.

at equilibrium
P=f+mg sinθ
f=Pmg sinθ
Now as P increases, f increases linearly w.r.t P.

Considering above mentioned points, if you consider up the plane +ve direction and down the plane as -ve direction, frictional force is represented by graph (a)

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